I need help in finding a force....with solutions please

Question

a pole vaulter carries a 6.0m long pole by pushing down with his right hand 0.40m from the right end and pushing up with his left hand, 1.0m from the same end. the pole is uniform and has a mass of 10.0kg. find the force exerted by his hand.

Answer 1

You are looking at the front of the person holding the pole.  

ASSUMPTIONS : the pole is horizontal.  And the pole is uniformly heavy throughout.

|==================|   10 Kg

......R

<--->

  0.4

...............L

<--------->

   1.0

<------------- 6 m ------------->

Upward force on the right  hand is that of the pole on the right hand side multiplied by the lever effect.  The left hand is the fulcrum.

Force x moment arm (right) = Force x moment arm (left)

Equation Fr x (1 - 0.4)  + [10 x (1/6) x 0.6/2] =  [10 x ( 5/6)]  x 5/2

On the left side of the equation above, the downwards moment has two components : the first is the downwards moment about the fulcrum of the right hand force, PLUS the downwards moment of the weight of the pole on that side.

The weight of the pole on the either side of the above diagram acts at 1/2 the length on that side, since it's a uniform pole.

Fr x 0.6   +   0.5  =  8.33 x 2.5  to two decimal places

Thus solving above; Fr = 34.21 Kg

Thus by a balance of forces, the left hand has to carry the load of the whole pole PLUS the right hand force as above.

Fl  =  10 + 34.21  =  44.21  Kg  to two decimal places.

That's your answer.

Pretty stupid way to carry the pole !